Given position vectors

$\overrightarrow{OA}=2\vec a, \qquad \overrightarrow{OB}=6\vec a+5\vec b, \qquad \overrightarrow{OC}=3\vec b$

Use the given area of a parallelogram

The parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ has area $|\overrightarrow{OA}\times \overrightarrow{OC}|=15$

Now, $\overrightarrow{OA}\times \overrightarrow{OC}=(2\vec a)\times(3\vec b)=6(\vec a\times \vec b)$

Hence, $6|\vec a\times \vec b|=15$ $|\vec a\times \vec b|=\frac{15}{6}=\frac{5}{2}$

Find the area of quadrilateral $OABC$

Split the quadrilateral into two triangles: $\text{Area}(OABC)=\text{Area}(\triangle OAB)+\text{Area}(\triangle OBC)$

Area of $\triangle OAB$

$\text{Area}(\triangle OAB)=\frac12\left|\overrightarrow{OA}\times \overrightarrow{OB}\right|$

Compute: $\overrightarrow{OA}\times \overrightarrow{OB}=(2\vec a)\times(6\vec a+5\vec b)$ $=12(\vec a\times \vec a)+10(\vec a\times \vec b)$ $=0+10(\vec a\times \vec b)$ $=10(\vec a\times \vec b)$

So, $\text{Area}(\triangle OAB)=\frac12\cdot 10|\vec a\times \vec b|=5\cdot \frac52=\frac{25}{2}$

Area of $\triangle OBC$

$\text{Area}(\triangle OBC)=\frac12\left|\overrightarrow{OB}\times \overrightarrow{OC}\right|$

Compute: $\overrightarrow{OB}\times \overrightarrow{OC}=(6\vec a+5\vec b)\times (3\vec b)$ $=18(\vec a\times \vec b)+15(\vec b\times \vec b)$ $=18(\vec a\times \vec b)+0$ $=18(\vec a\times \vec b)$

Therefore, $\text{Area}(\triangle OBC)=\frac12\cdot 18|\vec a\times \vec b|=9\cdot \frac52=\frac{45}{2}$

Total area

$\text{Area}(OABC)=\frac{25}{2}+\frac{45}{2}=\frac{70}{2}=35$

Check options

The required area is: $\boxed{35}$

So the correct option is C.